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      <a class="index-header" href="/michuanblog/2022/05/09/%E6%B1%82%E8%81%8C%E7%BB%8F%E9%AA%8C/">求职经验</a>

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        <div>计算机类（0809）求职经验简历
Keep Simple，极简主义，简历不需要华丽的字体和颜色，简单的黑白两色，保证全文字体一致，尽量一页纸

网站职徒简历-免费简历模板下载|中英文简历模板|智能求职简历工具
字体推荐微软雅黑，在电脑上显示效果最好
字号
标题小二号加粗
小标题小四号加粗
正文小四号

页数一页纸，尽量精简，简历写得越多，被考察的知识点就越多，选择自己掌握得比较熟悉的写。
内容校招</div>
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            <i class="iconfont icon-date"></i>&nbsp;2022-05-09
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      <a class="index-header" href="/michuanblog/2021/06/25/gaokao/">gaokao</a>

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        <div>高考志愿填报参考关于高考志愿填报的一些见解，有不足之处欢迎指正！
兴趣了解自己的兴趣，可能短短几天想弄清楚一个专业有困难，可以从几个方面入手：
（1）专业发展前景
（2）专业课程设置
（3）毕业工作领域
（4）毕业生就业情况
推荐在知乎、B站、学校本科招生官网、阳光高考网
查询相关信息。
推荐B站专业介绍视频：
B站专业选择指南工学篇
B站专业选择指南理学篇
B站专业选择指南经济学篇
城市定位城市</div>
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            <i class="iconfont icon-date"></i>&nbsp;2021-06-25
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      <a class="index-header" href="/michuanblog/2021/04/06/deep-learning-object-detection/">deep_learning_object_detection</a>

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        <div>deep learning object detectionAuthor: deep learning object detection
Paper list from 2014 to now(2019)The part highlighted with red characters means papers that i think “must-read”.However, it is my p</div>
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            <i class="iconfont icon-date"></i>&nbsp;2021-04-06
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      <a class="index-header" href="/michuanblog/2021/01/07/%E8%AF%BB%E4%B9%A6%E5%8D%95/">读书单</a>

      <a href="/michuanblog/2021/01/07/%E8%AF%BB%E4%B9%A6%E5%8D%95/" class="index-excerpt">
        
        
          
        
        <div>2019读书单
贾行家：《尘土》《潦草》
海桑：《不如让每天发生些小事》《我的身体里早已落叶纷飞》《我是你流浪过的一个地方》
惠特曼：《惠特曼诗选》
沈从文：《从文家书》
本哈德·施林克：《朗读者》
青山七惠：《温柔的叹息》
张枣：《张枣的诗》《春秋来信》
多多：《多多的诗》
海子：《海子的诗》
沈大成：《屡次想起的人》
王安忆 北岛选编：《给孩子的故事》
许知远《单向街002：先锋已死？》
双雪</div>
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            <i class="iconfont icon-date"></i>&nbsp;2021-01-07
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      <a class="index-header" href="/michuanblog/2020/09/13/lcof-68-2/">lcof_68_2</a>

      <a href="/michuanblog/2020/09/13/lcof-68-2/" class="index-excerpt">
        
        
          
        
        <div>LCOF68 - II 二叉树的最近公共祖先题目给定一个二叉树, 找到该树中两个指定节点的最近公共祖先。
解析若 root 是 p, q的 最近公共祖先 ，则只可能为以下情况之一：

p 和 q 在 root 的子树中，且分列 root 的 异侧（即分别在左、右子树中）；
p = root ，且 q 在 root 的左或右子树中；
q = root ，且 p 在 root 的左或右子树中；


考</div>
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          <div class="post-meta mr-3">
            <i class="iconfont icon-date"></i>&nbsp;2020-09-13
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              <a href="/michuanblog/categories/LeetCode-LCOF/">LeetCode_LCOF</a>
            
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              <a href="/michuanblog/tags/LeetCode/">LeetCode</a>
            
              <a href="/michuanblog/tags/C/">C++</a>
            
              <a href="/michuanblog/tags/LCOF/">LCOF</a>
            
              <a href="/michuanblog/tags/%E9%A2%98%E8%A7%A3/">题解</a>
            
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      <a class="index-header" href="/michuanblog/2020/09/13/lcof-68-1/">lcof_68_1</a>

      <a href="/michuanblog/2020/09/13/lcof-68-1/" class="index-excerpt">
        
        
          
        
        <div>LCOF68 - I 二叉搜索树的最近公共祖先题目给定一个二叉搜索树, 找到该树中两个指定节点的最近公共祖先。对于有根树 T 的两个结点 p、q，最近公共祖先表示为一个结点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）
解析
祖先的定义：若节点p在节点root的左（右）子树中，或p=root，则称root是p的祖先。
最近公共祖先的定义：设节点root为节</div>
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          <div class="post-meta mr-3">
            <i class="iconfont icon-date"></i>&nbsp;2020-09-13
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              <a href="/michuanblog/categories/LeetCode-LCOF/">LeetCode_LCOF</a>
            
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              <a href="/michuanblog/tags/LeetCode/">LeetCode</a>
            
              <a href="/michuanblog/tags/C/">C++</a>
            
              <a href="/michuanblog/tags/LCOF/">LCOF</a>
            
              <a href="/michuanblog/tags/%E9%A2%98%E8%A7%A3/">题解</a>
            
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      <a class="index-header" href="/michuanblog/2020/09/13/lcof-62/">lcof_62</a>

      <a href="/michuanblog/2020/09/13/lcof-62/" class="index-excerpt">
        
        
          
        
        <div>LCOF62 圆圈中最后剩下的数字题目0, 1, …, n-1这n个数字排成一个圆圈，从数字0开始，每次从这个圆圈里删除第m个数字。求出这个圆圈里剩下的最后一个数字。
解析正向推导过程: 约瑟夫环最后一个人的下标,一定是0(只剩一个人活着了),这点都能理解 所以从0推导

一个人的时候: 这个活着的人的下标是0. 所以需要知道当两个人存在的时候,这个人的下标是多少;
两个人的时候: 这个活着的人下</div>
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            <i class="iconfont icon-date"></i>&nbsp;2020-09-13
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              <a href="/michuanblog/categories/LeetCode-LCOF/">LeetCode_LCOF</a>
            
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              <a href="/michuanblog/tags/LeetCode/">LeetCode</a>
            
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      <a class="index-header" href="/michuanblog/2020/09/13/lcof-60/">lcof_60</a>

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        <div>LCOF60 n个骰子的点数题目把n个骰子扔在地上，所有骰子朝上一面的点数之和为s。输入n，打印出s的所有可能的值出现的概率。
需要用一个浮点数数组返回答案，其中第 i 个元素代表这 n 个骰子所能掷出的点数集合中第 i 小的那个的概率。
示例 2:
输入: 2
输出: [0.02778,0.05556,0.08333,0.11111,0.13889,0.16667,0.13889,0.11111</div>
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            <i class="iconfont icon-date"></i>&nbsp;2020-09-13
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              <a href="/michuanblog/categories/LeetCode-LCOF/">LeetCode_LCOF</a>
            
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              <a href="/michuanblog/tags/LeetCode/">LeetCode</a>
            
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      <a class="index-header" href="/michuanblog/2020/09/13/lcof-57-2/">lcof_57_2</a>

      <a href="/michuanblog/2020/09/13/lcof-57-2/" class="index-excerpt">
        
        
          
        
        <div>LCOF57 - II 和为s的连续正数序列题目输入一个正整数 target ，输出所有和为 target 的连续正整数序列（至少含有两个数）。
序列内的数字由小到大排列，不同序列按照首个数字从小到大排列。
解析三种方法：滑动窗口、求根公式、利用首尾元素间隔
代码1 滑动窗口vector&lt;vector&lt;int&gt;&gt; findContinuousSequence(int tar</div>
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            <i class="iconfont icon-date"></i>&nbsp;2020-09-13
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              <a href="/michuanblog/categories/LeetCode-LCOF/">LeetCode_LCOF</a>
            
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              <a href="/michuanblog/tags/LeetCode/">LeetCode</a>
            
              <a href="/michuanblog/tags/C/">C++</a>
            
              <a href="/michuanblog/tags/LCOF/">LCOF</a>
            
              <a href="/michuanblog/tags/%E9%A2%98%E8%A7%A3/">题解</a>
            
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      <a class="index-header" href="/michuanblog/2020/09/13/lcof-59-1/">lcof_59_1</a>

      <a href="/michuanblog/2020/09/13/lcof-59-1/" class="index-excerpt">
        
        
          
        
        <div>LCOF59 - I 滑动窗口的最大值题目给定一个数组 nums 和滑动窗口的大小 k，请找出所有滑动窗口里的最大值。
示例:
输入: nums = [1,3,-1,-3,5,3,6,7], 和 k = 3
输出: [3,3,5,5,6,7] 
解释: 

  滑动窗口的位置                最大值

[1  3  -1] -3  5  3  6  7       3
 1 [3  </div>
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            <i class="iconfont icon-date"></i>&nbsp;2020-09-13
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              <a href="/michuanblog/categories/LeetCode-LCOF/">LeetCode_LCOF</a>
            
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              <a href="/michuanblog/tags/LeetCode/">LeetCode</a>
            
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              <a href="/michuanblog/tags/LCOF/">LCOF</a>
            
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